Question

If a circle of radius 4 cm passes through the foci of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) and is concentric with the hyperbola, then the eccentricity of the conjugate hyperbola of that hyperbola is: 

• A. \( 2 \)
• B. \( 2\sqrt{3} \)
• C. \( \frac{2}{\sqrt{3}} \)
• D. \( \sqrt{3} \)

Hint:

For conjugate hyperbolas, remember that the eccentricity is calculated using the relation \( e' = \frac{c}{b} \). Also, knowing that the hyperbola's foci lie on the major axis helps in determining the parameters correctly.

Answer 1

The Correct Option is A

Step 1: Equation of the Given Hyperbola 
The given equation of the hyperbola is: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1. \] For a hyperbola, the foci are located at \( (\pm c, 0) \), where: \[ c^2 = a^2 + b^2. \] 
Step 2: Given Circle Passes Through Foci 
The circle is given to have a radius of 4 cm and passes through the foci. Since the center of the circle is the same as the hyperbola, we know that the distance from the center to the foci is equal to the circle's radius: \[ c = 4. \] Using \( c^2 = a^2 + b^2 \), we substitute \( c = 4 \): \[ 16 = a^2 + b^2. \] 
Step 3: Equation of Conjugate Hyperbola 
The conjugate hyperbola corresponding to the given hyperbola is: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1. \] For this hyperbola, the eccentricity \( e' \) is given by: \[ e' = \frac{c}{b}. \] Since \( c^2 = a^2 + b^2 \), we substitute \( c = 4 \): \[ e' = \frac{4}{b}. \] Since for a hyperbola, \( b^2 = a^2 - c^2 \), we use the identity: \[ b^2 = a^2 - (a^2 + b^2 - b^2) = 4. \] Thus, \( b = 2 \), and the eccentricity of the conjugate hyperbola is: \[ e' = \frac{4}{2} = 2. \] 
Final Answer: \( \boxed{2} \)

Answer 2

To solve this problem, we need to find the eccentricity of the conjugate hyperbola. Start by recalling that for a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the coordinates of the foci are \((\pm ae,0)\), where \( e \) is the eccentricity given by \( e = \sqrt{1+\frac{b^2}{a^2}} \). A circle passing through the foci is concentric with the hyperbola, thus, both are centered at the origin.

Given the circle’s radius is 4 cm and passes through the foci:

\( ae = 4 \)

Now, the conjugate hyperbola with equation \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) will have its eccentricity \( e' \) given by \( e' = \sqrt{1+\frac{a^2}{b^2}} \). Using the fact that \( b^2 = a^2(e^2-1) \) for the original hyperbola, substitute:

\( b^2 = a^2(\frac{16}{a^2} - 1) = 16 - a^2 \)

For the conjugate hyperbola:

\( e' = \sqrt{1+\frac{a^2}{16-a^2}} \)

Given the condition \( ae = 4 \), substitute \( e = \frac{4}{a} \) and \( e^2 = \frac{16}{a^2} \). Thus:

\( b^2 = a^2(\frac{16}{a^2} - 1) = 16 - a^2 \)

Apply this back to the conjugate hyperbola:

\( e' = \sqrt{1+\frac{a^2}{16-a^2}} = \sqrt{2} \cdot \frac{4}{a} = 2 \)

Thus, the eccentricity of the conjugate hyperbola is \( \boxed{2} \).